Distinct Odd-Town

Simple way to think this problem through:

First, choose the last digit. There are 5 options, since there are 5 odd numbers between 0-9.

Next, choose the first digit. There are 8 options, since 0 and your last digit are excluded.

Then, choose your middle digit. There are 8 options, since your other two digits are excluded.

5 * 8 * 8 = 320

We know, there are total $10$ digits possible in a face value, that are $1, 2, 3, 4, 5, 6, 7, 8, 9, 0.$

Now, to begin, we must first know all the conditions given in the question, that are: the number should be three digit , odd and all digits should be distinct .

Now, since the number is three digits, we have three places- Hundred's Place , Ten's Place and One's Place .

Since the number is odd, therefore, the possibilities of the one's digit are $1, 3, 5, 7, 9$ , therefore, $\fbox{5}$ possibilities.

Now, consider the hundred's digit. For the hundred's digit, possibilities are $1, 2, 3, 4, 5, 6, 7, 8, 9, 0$ , but since the number has to be a $3$ digit number, $0$ can't be a possibility, also, one of the remaining $9$ numbers would be chosen by the one's place, so we have to remove one more possibility in order to follow the condition of distinct digits, leaving us with $\fbox{8}$ possibilities for the hundred's place.

Now, for the ten's place, we again shall have $10$ possibilities, but we must eliminate two of them, because two distinct digits must be occupied by the hundred's and the one's place, leaving us with $\fbox{8}$ possibilities for the ten's place.

Therefore, we have possibilities of $8, 8$ and $5$ for the distinct digits, and thus we can apply the Rule Of Product to get the total number of possibilities of the numbers which satisfy the given conditions, giving us $8 \times 8\times 5 = 320$ .

Therefore, the maximum number of houses in Odd Town is $\fbox{320}$ .

Though the initial wording of the problem may seem a bit confusing, all the problem is asking for is the number of odd three digit numbers that have distinct digits.

We know that the units digit must be odd, and the odd digits are 1, 3, 5, 7, and 9, for a total of five choices for the units digit.

The hundreds digit cannot be 0 (else it is not a 3-digit number) or the equivalent of the units digit (we cannot have a repeated digit), which leaves a total of $10-1-1=8$ possible values.

Finally, the tens digit cannot be the same as the units digit or the hundreds digit, for a total of 8 possible values. (Note that the tens digit can be equal to 0.)

Multiplying, we have $5\cdot8\cdot8=\boxed{320}$ .

Since the houses are labelled with 3-digit positive integers that are odd which are all distinct, then we must realize that these integers can be subdivided into 4 possible cases:

• Hundreds digit is Odd ; Tens digit is Even ; Ones digit is Odd
• Hundreds digit is Even ; Tens digit is Odd ; Ones digit is Odd
• Hundreds digit is Even ; Tens digit is Even ; Ones digit is Odd
• Hundreds digit is Odd ; Tens digit is Odd ; Ones digit is Odd

In the first case, we must consider 5 possible choices for Ones digit (which are 1,3,5,7,9). Then for the Hundreds digit, there are only 4 choices left because we have chosen one odd number for the Ones digit. Then for the Tens digit, there are 5 possible choices (which are 0,2,4,6,8). So, multiply all the possible choices for Hundreds, Tens, and Ones then we have $4 \times 5 \times 5 = 100$

In the second case, we must again consider 5 choices for the Ones digit. Then we must consider 4 possible choices for the Hundreds digit (which are 2,4,6,8). Then we consider 4 choices left for the Tens digit since we have already chosen one odd number for the Ones digit. So, $4 \times 4 \times 5$ will give us $80$ .

In the third case, consider 5 choices for Ones digit. Then 4 choices for the Hundreds digit and 4 choices left for the Tens digit since we have already chosen one even number for the Hundreds digit but in this case, 0 is included in our choice. So, $4 \times 4 \times 5 = 80$ .

In the last case, we consider 5 choices for the Ones digit again. Then there are 4 choices left for the Hundreds digit and 3 choices left for the tens digit. So, $4 \times 3 \times 5 = 60$ .

$100 + 80 + 80 + 60 = \boxed{320}$

Hence, there are 320 houses as the maximum possible number of houses in odd-town.

By using filling slot, we have 3 slots and ten potential number to fill (0 to 9) . Fill the units possibility, there should be 5 possibilities, namely: 1, 3, 5, 7, and 9 . The first digit has 8 possibilities since 0 and the units digit are excluded. Then, the tens digit is left with 8 another choices since 0 is allowed and two possibilities were used by the units and hundreds digit. Thus, __8 x 8 x 5 = \boxed{320}) is the answer.

There are 5 possibilities for the unit digit of an odd number: $1,3,5,7,9$ .

$5\times P(9,2)=5\times9\times8=360$

However we need to minus $8\times5=40$ from $360$ ,because the digit number begin with digit $0$ does not exist.

Ans= $360-40=320$

case1 0 at tenth digit,8 possibliities in hundredth digit ,5 in ones digit=total =40 case2 5 in unit digit,8 in tenth,7 in thousand=5 8 7=280 so 40+280=320...........

Lets divide the three digit numbers into three places. At hundreds place we have 8 choices of numbers(i,e excluding 0 and odd number at ones place). At tens place we again have 8 choices (i.e excluding number at hundreds place and ones place as now we can put zero in tens place) and in ones place we have 5 choices that are the odd numbers. Hence ans = 8 X 8 X 5 = 320

the 3 digits can be as follows: (E = even digit) and (O=odd digit)

EEO , EOO , OEO , OOO

also note that in the hundreds place 0 can not come

so now possibilities for

EEO = 4 4 5 = 80

EOO = 4 5 4 = 80

OEO = 5 5 4 = 100

OOO = 5 4 3 = 60

therefore total possibilities is 80+80+100+60 = 320

ALLITERATE METHOD :

1) In the one's place we can put any 5 odd numbers

2) In the hundreds place we can put 8 numbers (because the digit in one's place and zero are excluded)

3) In ten's place we can put 8 numbers (because the digits in ones place and ten's place are excluded)

therefore total is 8 8 5 = 320

There are

(8 x 7 x 1 )+ (8 x 1 x 1 ) = 64 ways of choosing 3-digit odd number with 1 as the last digit.

(8 x 7 x 1 )+ (8 x 1 x 1 ) = 64 ways of choosing 3-digit odd number with 3 as the last digit.

(8 x 7 x 1 )+ (8 x 1 x 1 ) = 64 ways of choosing 3-digit odd number with 5 as the last digit.

(8 x 7 x 1 )+ (8 x 1 x 1 ) = 64 ways of choosing 3-digit odd number with 7 as the last digit.

(8 x 7 x 1 )+ (8 x 1 x 1 ) = 64 ways of choosing 3-digit odd number with 9 as the last digit.

So we have , 64 (5) = 320 ways in choosing a 3-digit odd number with distinct digits. Hence there are 320 maximum houses in the odd-town.